If g(x) = 3x + sqrt x

[diananah]

If g(x) = 3x + sq rt of x, what is the value of g(d^2 +6d+9)?
(Chapter 5, Page 79 - In Action problem #2)

I'm struggling with the rule: sq roots have one value. In this problem, the answer key lists two solutions: one which has d+3 is greater than/= zero and the other where d+3 is less than/= zero. I'm unclear as to why this problem doesn't have just one answer - where d+3 greater than/= zero, given the general rule that when we see a sq root, we should assume it to be the positive value.

[mithunsam]

We can write g(d^2 + 6d + 9) as g((d+3)^2)

But, g(x) = 3x + sqrt(x)
Therefore, g((d+3)^2) = 3((d+3)^2) + sqrt((d+3)^2)
=> 3((d+3)^2) + (d+3) or ((d+3)^2) -(d+3)
=> 3(d^2 + 6d + 9) + (d + 3) or 3(d^2 + 6d + 9) - (d + 3)
=> 3d^2 + 18d + 27 + d + 3 or 3d^2 + 18d + 27 -d - 3
=> 3d^2 + 19d + 30 or 3d^2 + 17d + 24

It should be the solution, please verify.

Usually, we take only +ve values for the sqrt of a variable. However, in this special case, we aren't taking sqrt of a variable. We are taking sqrt of the square of a variable.

That is, sqrt((d+3)^2)

For both +ve and -ve values of (d+3), (d+3)^2 will be +ve. That means, we have to take square roots for two +ve values - one is (d+3)^2, when (d+3) is +ve and the second one is (d+3)^2, when (d+3) is -ve. So, sqrt((d+3)^2) will have two +ve solutions.

We shouldn't ignore any +ve solutions. Therefore, the problem has two solutions.

[diananah]

That was really helpful, thanks so much!

[tim]

cool!

[sanmayved]

I understood mithunsam's explanation but then page 98 in the same book contradicts this by saying:

Sqrt x^2 = |x|

So which one do we follow?

[tim]

there is no contradiction here. sqrt(x^2) does in fact equal |x|. one way you can deal with the absolute value is to take a positive and negative version of what's inside, subject to whatever other constraints are in the problem. what both mithunsam and our book failed to do is to note that each of these answers is only valid for a certain range of possible x values..